Question 4832 – Operating-Systems
November 17, 2023Question 4999 – Operating-Systems
November 17, 2023Question 4834 – Operating-Systems
For the implementation of a paging scheme, suppose the average process size be ‘x’ bytes, the page size be ‘y’ bytes, and each page entry requires ‘z’ bytes. The optimum page size that minimizes the total overhead due to the page table and the internal fragmentation loss is given by
Correct Answer: C
Question 602 Explanation:
Since the average number of pages required per process will be x/y and the amount of space required by the page table will be (x/y)*z. The amount of space lost due to internal fragmentation is y/2. So total space wastage is
Loss(L)=(x/y)*e + y/2
To find the value of ‘y’ that yields the minimal values, take rst derivative with respect to ‘y’ and set the resulting equation to zero.
(dL/dy) =0
y=√(2xz)
Loss(L)=(x/y)*e + y/2
To find the value of ‘y’ that yields the minimal values, take rst derivative with respect to ‘y’ and set the resulting equation to zero.
(dL/dy) =0
y=√(2xz)
x/2
xz/2
√2xz
√ xz/ 2
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